Question

我正在用Javascript创建一个二十一点纸牌游戏。到目前为止，我感谢你的评论。我正在寻找反馈，为什么我只能在cardDealer函数中调用cardFace函数一次。目前cardFace（0）给了我一个值，而cardFace（1）或任何其他数字都没有，除非它是cardDealer函数中唯一被调用的cardFace（x）。任何想法，将不胜感激。谢谢

下面的代码并不完整，但我相信足以从中辨别出足够的信息。谢谢

    function card(name, suit, face, value)
    {
        this.name = name;
        this.suit = suit;
        this.face = face;
        this.value = value;
    }

    aceOfHearts = new card("aceofhearts","hearts", "ace", 11);
    kingOfHearts = new card("kingofhearts","hearts", "king", 10);
    queenOfHearts = new card("queenofhearts","hearts", "queen", 10);
    jackOfHearts = new card("jackofhearts","hearts", "jack", 10);
    tenOfHearts = new card("tenofhearts","hearts", "ten", 10);
    nineOfHearts = new card("nineofhearts","hearts", "nine", 9);
    eightOfHearts = new card("eightofhearts","hearts", "eight", 8);
    sevenOfHearts = new card("sevenofhearts","hearts", "seven", 7);
    sixOfHearts = new card("sixofhearts","hearts", "six", 6);


    var deck;
    deck = [];


    deck.push(aceOfHearts);
    deck.push(aceOfDiamonds);
    deck.push(aceOfClubs);
    deck.push(aceOfSpades);
    deck.push(kingOfClubs);
    deck.push(kingOfDiamonds);
    deck.push(kingOfHearts);
    deck.push(kingOfSpades);
    deck.push(queenOfClubs);
    deck.push(queenOfDiamonds);
    deck.push(queenOfHearts);
    deck.push(queenOfSpades);
    deck.push(jackOfClubs);
    deck.push(jackOfDiamonds);
    deck.push(jackOfHearts);
    deck.push(jackOfSpades);
    deck.push(tenOfClubs);
    deck.push(tenOfDiamonds);
    deck.push(tenOfHearts);
    deck.push(tenOfSpades);


    var cardDealer = function()
    {  
        fisherYates(deck);

        document.getElementById("yourFirstCard").textContent = "Your First Card is " + cardFace(0) ;
        document.getElementById("yourSecondCard").textContent = "Your Second Card is " + cardFace(1) ;

    };

    var cardFace = function(x)
    {   
        cardFace = deck[x].face;
        return cardFace;
    };


    var cardSuit = function(x)
    {
        cardSuit = deck[x].suit;
        return cardSuit;
    };

    <body>
<div>Black Jack 1.1</div>
<br/>
<button type="button" onClick="cardDealer()">Deal</button> 
<div id="yourFirstCard"></div>
<div id="yourSecondCard"></div>
<div id="yourThirdCard"></div>
<div id="yourFourthCard"></div>
<div id="yourFifthCard"></div>
<div id="playerTotal"></div></div>
<div id="playerSit"></div>
<div id="playerMessage"></div>
<div id="sitter"><button type="button" onClick="sit()">Sit</button></div>
<div id="hitter"><button type="button" onClick="hitMe()">Hit</button></div> 
<br/>
<br/>
<div id="dealersFirstCard"></div></div>
<div id="dealersSecondCard"></div></div>
<div id="dealersThirdCard"></div></div>
<div id="dealersFourthCard"></div></div>
<div id="dealersFifthCard"></div></div>
<div id="dealerTotal"></div></div>
<div id="dealerSit"></div></div>
<div id="dealerMessage"></div></div>
</body>

Answer 1

以下是我看到问题的地方：

var cardFace = function(x)
{   
    cardFace = deck[x].face;
    return cardFace;
};

首先，声明一个名为cardFace的变量并将其分配给一个函数。实际上这个函数重新分配了你的cardFace变量并将其作为结果返回。就是这样，你的功能现在已经消失，你不能再引用它，因为cardFace已被重新分配。

您需要做的是更改函数的主体，使其不重新分配变量：

var cardFace = function(x)
{   
    var cardFaceValue = deck[x].face;
    return cardFaceValue;
};

cardSuit功能也是如此。

Answer 2

将cardFace移到cardDealer之上或将其更改为function cardFace(x)定义。

这是一个js行为的事情。以function x格式定义的函数可以从上面编写的代码中调用。 var x = function()无法在运行时确定的位置。

请参阅：http://ejohn.org/apps/learn/#4了解Resig先生的更多示例。

试试：http://ejohn.org/apps/learn/这是一件好事。

编辑：在这个帖子中你也应该考虑很多好的建议。包括其他发布的答案。你应该研究scopes。

编辑2

在再次查看您的代码之后，我相信我的解决方案实际上并不是正确的，因为它让我想到cardFace在cardDealer的范围内被调用，这意味着它可能是由您致电cardDealer的时间定义。因此，我认为当你将其改为function时，你可能会改变名称，正如另一个答案所暗示的那样。而且实际上是正确的答案。

//Test function calling forward defined function
var f = function() { x(); };
var x = function() { console.log('a'); }
f()
>>> a

// Test function modifying it's own scoped name
var f = function() { f = 'test'; return f; };
f()
>>> "test"
f()
>>> TypeError: string is not a function

// Test function modifying it's own scoped name when declaring as function
function f() { f = 'test'; return f; };
f()
>>> "test"
f()
>>> TypeError: string is not a function

Javascript ...显示值

2 个答案: